IB Chemistry on Entropy, Gibbs Free Energy, Second Law of Thermodynamics
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IB Chemistry on Entropy, Gibbs Free Energy, Second Law of Thermodynamics

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IB Chemistry on Entropy, Gibbs Free Energy, Second Law Thermodynamics and Spontaneity of reaction

IB Chemistry on Entropy, Gibbs Free Energy, Second Law Thermodynamics and Spontaneity of reaction

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IB Chemistry on Entropy, Gibbs Free Energy, Second Law of Thermodynamics IB Chemistry on Entropy, Gibbs Free Energy, Second Law of Thermodynamics Presentation Transcript

  • Entropy, Free Energy and Spontaneity Entropy• Measures the degree of DISORDER for a system• Measures the probability or chance for a system, both in distribution of particles in space and distribution of energy • How will a reaction go ? • Why gas MIXES and NOT UNMIX ? • Why concentrated solute DIFFUSE and NOT UNDIFFUSE?
  • Entropy and SpontaneityImportant concepts for Entropy• Unit for entropy – JK-1mol-1• Formula for entropy, S = k ln W k = Boltzmann constant W = Number ways a particle can arrange in space• Absolute entropy can be measured!• Entropy of a perfectly order solid crystal at OK is ZERO• Entropy of an element under standard condition is NOT ZERO ΔSθ (H2) gas = 130.6JK-1mol-1• Standard entropy change, ΔSθ = Entropy change per mole of a substance heated from OK to standard Temp of 298K
  • Entropy and SpontaneityENTROPY CHANGE, ΔSθ• Change in the disorder of a system• Higher ↑ disorder – Higher entropy ↑• Change in state from SOLID → LIQUID → GAS → Higher entropy ↑• Greater ↑ number of particles formed in products → Higher entropy ↑• Complex molecules (more atoms bonded) → Higher entropy ↑• Higher temperature ↑ → Particles vibrate faster → More random → Higher entropy ↑Standard entropy, ΔSθ /298K Sθ / JK-1mol-1
  • Second Law of Thermodynamics• For all spontaneous reactions, TOTAL entropy of the universe ΔSθuni always INCREASES • Formula for ΔSθsys = ΔSθ (products) - ΔSθ (reactants) • Formula for ΔSθ(surrounding) • Reaction will likely to happen if - • Conclusion : For spontaneous reaction, ΔSθuni must > O (positive)
  • Gibbs Free Energy, ΔG To predict if a reaction will likely to happen •ΔHsys = -ve (exothermic) • ΔSsys = +ve (entropy increses ↑) • Δssurr = +ve (entropy increases ↑ ) •ΔSuni = +ve (entropy increases ↑ ) • ΔSuni = ΔSsys + ΔSsurr > 0, (+ve) positive • ΔGsys =ΔHsys –TΔSsys < 0, (-ve) negativeGibbs Free Energy, ΔG is used• ΔG is used or preferred because it involves the system while ΔS involves system and surrounding• Easier to determine ΔH and ΔS for a system ΔG = ΔH -TΔS Unit for ΔGsys = kJmol-1 ΔHsys = kJmol-1 ΔSsys = JK-1mol-1
  • Combination ΔHsys , ΔSsys and ΔGsys to predict if reaction is spontaneous ΔHsys -ve ΔSsys +ve Always Spontaneous ΔHsys +ve ΔSsys –ve Non Spontaneous ΔHsys +ve ΔSsys +ve ↑ Spontaneous if Temp High ↑ ΔHsys -ve ΔSsys –ve ↓ Spontaneous if Temp low ↓
  • Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K ΔSθsys ΔSθsurrΔSsys = ΔSsys (product) – ΔSsys (reactant) ΔSsurr = - ΔHsys /T = (70.0) – (188.7) + = -(-44100)/298 = -118.7 JK-1mol-1 = +148.0JK-1mol-1 ΔSuni = ΔSsys + ΔSsurr = (-118.7) + 148.0 = +29.3 JK-1mol-1 AND ΔGθ = ΔH -TΔS = (-44.1) – (298 x -118.7/1000) = - 8.72kJmol-1Conclusion :• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ BUT heat released to the surrounding INCREASES ↑ the entropy of surrounding, ΔSsurr• Spontaneous - ΔGθ = -ve
  • Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K ΔSθsys ΔSθsurrΔSsys = ΔSsys (product) – ΔSsys (reactant) ΔSsurr = -ΔHsys /T = (213.6 + 2x171) – (186.0 + 2x205) + = -(-890000)/298 = -40.4 JK-1mol-1 = +2986.5JK-1mol-1 ΔSuni = ΔSsys + ΔSsurr = -40.4 + 2986.5 = +2946 JK-1mol-1 AND ΔGθ = ΔH -TΔS = (-890) – (298 x -40.4/1000) = - 878 kJmol-1Conclusion :• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ BUT heat released to the surrounding INCREASES ↑ the entropy of surrounding, ΔSsurr• Spontaneous - ΔGθ = -ve
  • Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K ΔSθsys ΔSθsurrΔSsys = ΔSsys (product) – ΔSsys (reactant) ΔSsurr = -ΔHsys /T = (2 x 115) – (130.6) + = -(+436000)/298 = +99.4 JK-1mol-1 = -1463 JK-1mol-1 ΔSuni = ΔSsys + ΔSsurr = +99.4 + (-1463) = -1363.6 JK-1mol-1 AND ΔGθ = ΔH -TΔS = (436) – (298 x 99.4/1000) = +406 kJmol-1Conclusion :• Non Spontaneous - Entropy of system ΔSθsys INCREASES ↑ BUT heat absorbed from surrounding DECREASES ↓ the entropy of surrounding, ΔS surr• NON spontaneous - ΔGθ = +ve
  • Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K Is it possible to freeze water at 298K ( 25oC)At 25oC ΔSθsys ΔSθsurrΔSsys = ΔSsys (product) – ΔSsys (reactant) ΔSsurr = -ΔHsys /T = (48) – (70) + = -(-6010)/298 = -22 JK-1mol-1 = +2016 JK-1mol-1 ΔSuni = ΔSsys + ΔSsurr = -22 + (+2016) = -1.84 JK-1mol-1 AND ΔGθ = ΔH -TΔS = (-6.01) – (298 x -22/1000) = +0.54 kJmol-1Conclusion :• Non Spontaneous - Entropy of system ΔSθsys DECREASES ↓ more than > the INCREASE ↑ in entropy of surr, ΔSsurr due to heat released to the surrounding• NON spontaneous - ΔGθ = +ve
  • Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 263K Is it possible to freeze water at 263K ( -10oC)At -10oC ΔSθsys ΔSθsurrΔSsys = ΔSsys (product) – ΔSsys (reactant) ΔSsurr = -ΔHsys /T = (48) – (70) + = -(-6010)/263 = -22 JK-1mol-1 = +22.85 JK-1mol-1 ΔSuni = ΔSsys + ΔSsurr = -22 + (+22.85) = +0.85 JK-1mol-1 AND ΔGθ = ΔH -TΔS = (-6.01) – (263 x -22/1000) = -0.23 kJmol-1Conclusion :• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ BUT heat released to surrounding INCREASES ↑ the entropy of surrounding, ΔSsurr• Spontaneous - ΔGθ = -ve
  • Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K Is it possible to decompose CaCO3 (s) → CaO (s) + CO2 (g) At 25oC ΔSθsys ΔSθsurrΔSsys = ΔSsys (product) – ΔSsys (reactant) ΔSsurr = -ΔHsys /T = (39.7 + 213.6) – (92.9) + = -(+178300)/298 = + 160.4 JK-1mol-1 = - 598.3 JK-1mol-1 ΔSuni = ΔSsys + ΔSsurr = +160.4 + (- 598.3) = - 438 JK-1mol-1 AND ΔGθ = ΔH -TΔS = (+ 178.3) – (298 x +160.4/1000) = + 130.5 kJmol-1Conclusion :• Non Spontaneous - Entropy of system ΔSθsys INCREASES ↑ BUT heat absorbed from surrounding DECREASES ↓ the entropy of surrounding, ΔSsurr• NON spontaneous- ΔGθ = +ve
  • Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 1500K Is it possible to decompose CaCO3 (s) → CaO (s) + CO2 (g)At 1227oC ΔSθsys ΔSθsurr ΔSsys = ΔSsys (product) – ΔSsys (reactant) ΔSsurr = -ΔHsys /T = (39.7 + 213.6) – (92.9) + = -(+178300)/1500 = + 160.4 JK-1mol-1 = - 118.8 JK-1mol-1 ΔSuni = ΔSsys + ΔSsurr = +160.4 + (- 118.8) = + 41.6 JK-1mol-1 AND ΔGθ = ΔH -TΔS = (+ 178.3) – (1500 x +160.4/1000) = - 62.3 kJmol-1 Conclusion : • Spontaneous - Entropy of system ΔSθsys INCREASES ↑ more than > DECREASES ↓ in entropy of surr, ΔSsurr due to heat absorbed from surrounding • Spontaneous - ΔGθ = -ve
  • Combination ΔH, ΔS and ΔG to predict if reaction is spontaneous at 298K Is the reaction possible 2NO (g) + O2 (g) → 2NO2 (g) ΔSθsys ΔSθsurrΔSsys = ΔSsys (product) – ΔSsys (reactant) ΔSsurr = -ΔHsys /T = (2 x 240) – (2 x 210.7+102.5) + = -(-114000)/298 = - 43.9 JK-1mol-1 = +382.5 JK-1mol-1 ΔSuni = ΔSsys + ΔSsurr = -43.9 + (+382.5) = +339 JK-1mol-1 AND ΔGθ = ΔH -TΔS = (- 114) – (298 x -43.9/1000) = -100.9 kJmol-1Conclusion :• Spontaneous - Entropy of system ΔSθsys DECREASES ↓ less than < INCREASE ↑ in entropy of surr, ΔSsurr due to heat released to the surrounding• Spontaneous - ΔGθ = -ve
  • Reaction which is Temperature Dependent whereby ΔHsys (+ve) and ΔSsys (+ve) Finding the temperature which make reaction spontaneous ΔGsys = (-ve) Spontaneous ΔGsys = (+ve) Non Spontaneous ΔGsys = O (Equilibrium)Equilibrium Temperature whereby reaction becomes spontaneous happen when ΔG = O ΔG = ΔHsys - TΔSsys 0 = ΔHsys -TΔSsys TΔS = ΔH T = ΔH/ΔS At what temperature will reaction becomes spontaneous ΔHsys = +178.3 kJmol-1 ΔSsys = +160.4 JK-1mol-1 0 = ΔH -TΔS TΔS = ΔH T = ΔH/ΔS T = 178300/160.4 = 1111.6K
  • Standard Enthalpy Change of Formation, ΔHf• Energy released when ONE mole of compound form from its constituent elements in their std states EX : C(s) + 2H2 → CH4 (g) ΔHf = -75 kJmol-1• Using standard enthalpy of formation to calculate ΔH reactionStandard Free Energy of Formation, ΔGf• Free energy released when ONE mole of compound form from its constituent elements in their std states EX : C(s) + 2H2 → CH4 (g) ΔGf = -51 kJmol-1• Using standard free energy of formation to calculate ΔG reaction ΔGreaction = (Sum ΔGf (product) – Sum ΔGf(reactant) ) ΔGreaction = (-ve) Reaction is Spontaneous• Standard free energy of formation, ΔGf of compound C(s) + 2H2 → CH4(g) ΔGf = -51 kJmol-1 H2(g) + ½ O2(g) →H20(g) ΔGf = -228 kJmol-1 2C (s) + 3H2(g) → C2H6(g) ΔGf = -33 kJmol-1• Standard free energy of formation, ΔGf of element is ZERO C(s) → C(s) ΔGf = 0 kJmol-1
  • Using ΔG to predict if reaction is SpontaneousEx 1 ΔGrxn = (Sum ΔGf (product) – Sum ΔGf (reactant) ) = ( -394 + 2 x -237) – (-51 + 0) = - 817 kJmol-1 (Spontaneous)Ex 2 ΔGrxn = (Sum ΔGf (product) – Sum ΔGf (reactant) ) = ( 3 x -303.2 + (-409.2) ) – ( 4 x -296.3) = -134 kJmol-1 (Spontaneous)Ex 3 ΔGrxn = (Sum ΔGf (product) – Sum ΔGf (reactant) ) = (-305) – (-267.8 + 0 ) = -37.2 kJmol-1 (Spontaneous)
  • Combination ΔH, ΔS and ΔG to predict if reaction is SpontaneousEx 1 ΔHrxn = (Sum ΔHf (product) – Sum ΔHf (reactant) ) = ( -986.6) – (-635.5 + (-285.9) ) ΔGθ = ΔH -TΔS = - 65.2 kJmol-1 = (-65.2) – (298 x -33.6/1000) ΔSrxn = (Sum S(product) – Sum S(reactant) ) = -55.2 kJmol-1 = ( +76.1) – (+39.7 + 70.0) (Spontaneous) = - 33.6 JK-1mol-1Ex 2 ΔHrxn = (Sum ΔHf (product) – Sum ΔHf (reactant) ) = ( 3 x -432.8 -436.7) – ( 4 x -397.7) ΔGθ = ΔH -TΔS = - 144 kJmol-1 = (-144) – (298 x -36.8/1000) ΔSrxn = (Sum S(product) – Sum S(reactant) ) = -133 kJmol-1 = ( 3 x 151 + 82.6) – ( 4 x 143.1) = - 36.8 JK-1mol-1 (Spontaneous)